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Has anyone else noticed that "Monoid" suggests, single or alone and this seems to pervade his whole persona, with his reluctance and difficulty to talk with and relate to people in real life? This isn't a jibe at the chap, I like him, I was just making an observation. |
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A set, M, defined on rational vectors, such that: (1) 0 is in M, and (2) if v, w are in M, then v+w is in M. The monoid is integral if it contains only integer vectors. One can think of a monoid as a discrete analogue of a convex cone. |
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euhm....yes! |
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Monoid |
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moonid |
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Monoid = Freddy Prince Jr. |
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Monoid homomorphisms A homomorphism between two monoids (M, *) and (M′, @) is a function f : M → M′ such that * f(x*y) = f(x)@f(y) for all x, y in M * f(e) = e′ where e and e′ are the identities on M and M′ respectively. Not every magma homomorphism is a monoid homomorphism since it may not preserve the identity. Contrast this with the case of group homomorphisms: the axioms of group theory ensure that every magma homomorphism between groups preserves the identity. For monoids this isn't always true and it is necessary to state it as a separate requirement. A bijective monoid homomorphism is called a monoid isomorphism. Two monoids are said to be isomorphic if there is an isomorphism between them. |
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mono id single identity |
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it always reminds me of hemorrhoid. |
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Not every monoid sits inside a group. For instance, it is perfectly possible to have a monoid in which two elements a and b exist such that a*b = a holds even though b is not the identity element. Such a monoid cannot be embedded in a group, because in the group we could multiply both sides with the inverse of a and would get that b = e, which isn't true. A monoid (M,*) has the cancellation property (or is cancellative) if for all a, b and c in M, a*b = a*c always implies b = c and b*a = c*a always implies b = c. A commutative monoid with the cancellation property can always be embedded in a group. That's how the integers (a group with operation +) are constructed from the natural numbers (a commutative monoid with operation + and cancellation property). However, a non-commutative cancellative monoid need not be embeddable in a group. If a monoid has the cancellation property and is finite, then it is in fact a group. An inverse monoid, is a monoid where for every a in M, there exists a unique a-1 in M such that a=aa-1a and a-1=a-1aa-1. |
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a cluster of monoids groped at her legs. |
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don't you mean that a semigroup of monoids with an identity element groped at her legs? |
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no. the whole idea of using the word "identity" with the word "monoid" rings false to me. |
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so the fact that any semigroup S may be turned into a monoid simply by adjoining an element e not in S and defining ee = e and es = s = se for all s ∈ S, which can be done recursively (i.e. if S is monoid with identity e, one obtains a new monoid S ∪ {e′} with identity e′, and so on) isn't a possibility? |
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p!ouow |
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y'all niggaz need to put the crack pipe down... |
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i'll put it down when i'm finished homie |
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? |
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HaHaH! |
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I always thought it was for monotonous. |
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Please do not create nonsense articles such a Phobiazero. Thank you. -- BMIComp (talk, HOWS MY DRIVING) 17:12, 2 September 2005 (UTC) This is the discussion page for an anonymous user who has not created an account yet or who does not use it. We therefore have to use the numerical IP address to identify him/her. Such an IP address can be shared by several users. If you are an anonymous user and feel that irrelevant comments have been directed at you, please create an account or log in to avoid future confusion with other anonymous users |
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Hey Drunken, are you a mastahmatician? |
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hahah! my foot fell asleep, so no. |
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hes a drunkenanything you want |
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so, where all of those smart lines come from? booze? |
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you betcha! booze and wikipedia. |
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true, thoes two makes a hell of a combination for smartness. though, wikis sometimes unnecessary. booze'll do. |
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booze booze booze and snooze! |
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and a bruise and a cruise and a lose and a muse |
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and being loose |
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you guys are right! |
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don't you mean "tight," you naughty panda? |
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hahaha! I like rum! :D |
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RUM! Cuba libre motherfucker!!! |
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yeah, bacardi means party!!! |
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:D!!!!!!!!! |
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